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Colt1911 [192]
3 years ago
5

Suppose that the instruction format for a modified Little Man Computer requires two consecutive locations for each instruction.

The high-order digits of the instruction are located in the first mail slot, followed by the low –order digits. The IR is large enough to hold the entire instruction and can be addressed as IR [high] and [IR] low to load it. You may assume that the op code part of the instruction uses IR [high] and that the address is found in IR [low]. Write the fetch-execute cycle for an ADD instruction on this machine.
Computers and Technology
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

// Fetch-execute cycle for an ADD instruction

PC →MAR

MDR → IR [high]

PC + 1 → PC

PC →MAR

MDR → IR [low]

IR [low] → MAR

MDR + A → A

PC + 1 → PC

Explanation:

  • MAR  is assigned PC and IR[high] value is assigned to   MDR.
  • Increment value of PC by 1 .
  • Again MAR  is assigned PC  and A is assigned MDR.
  • At the end, increment value of PC by 1 .
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A(n) _______ is the most basic type of access query
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I hope this helps! :)
4 0
3 years ago
Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
3 years ago
Which type of navigation involves multiple frames that are linked to a number of other frames?
sesenic [268]

Answer:

1. b)

2. c)

3. c)

4. a)

5. b)

Explanation:

1. and 5. Linear kind of navigation is a system with a sequential manner web pages that are perfect for some sorts of sites that are having information that has to be viewed as a book (5) and when we are talking about that view we are considering one page after another page like we are reading a book. It is also the simplest navigation. This is the explanation for question 1 and question 5.

2. The most well-designed navigation system is an intuitive one because in this design of the website we have website traffic that is easy because it flows from one web page to another web page. It is showing us where to go to find and look for something and even where to go if there is no concrete options for what are we looking for.

3. A Sitemap is referring to the organized hierarchy of links and it is the protocol that is allowing us to search through many links. A Sitemap is having a listing of the URLs for some site and that is why this is the correct answer.

4. In using liner reciprocal navigation the interface should include how frames are left and how many of them are there. The more the frames, the more times the user will spend on them and the site.

8 0
3 years ago
In this lab, you complete a C++ program that swaps values stored in three int variables and determines maximum and minimum value
Sergio039 [100]

Answer:

Following are the code to the given question:

#include <iostream>//header file

using namespace std;

int main()//main method

{

int first = 0,second = 0,third = 0;//defining integer variable  

int temp; //defining integer variable

const string SENTINEL = "done"; // defining a string variable as constant  

string repeat;// defining a string variable  

bool notDone = true; //defining bool variable

cout << "Enter first number: ";//print message

cin >> first;//input value

cout << "Enter second number: ";//print message

cin >> second;//input value

cout << "Enter third number: ";//print message

cin >> third;//input value

while(notDone == true)//defining a loop to check the value  

{

if(first > second)//use if to compare first and second value

{

int temp = first;//defining temp to hold first value

first = second;//holding second value in first variable

second = temp;//holding temp value in second variable

}

if(second > third)//use if to compare second and third value

{

int temp = second;//defining temp to hold second value

second = third;//holding second value in third variable

third = temp;//holding temp value in third variable

}

cout << "Smallest: " << first << endl;//print smallest value

cout << "Next smallest: " << second << endl;//print Next smallest value

cout << "Largest: " << third << endl;////print Largest value

cout << "Enter any letter to continue or done to quit: ";//print message

cin >> repeat;//holding string value

if (repeat == SENTINEL)

{

notDone = false;//holding bool value

}  

else //else block

{

cout << "Enter first number: ";//print message

cin >> first;//input value

cout << "Enter second number: ";//print message

cin >> second;//input value

cout << "Enter third number: ";//print message

cin >> third;//input value

}

return 0;

}

}

Output:

Please find the attached file.

Explanation:

  • Inside the main method Four integer variable "first, second, third, and temp" is declared in which first three variable is used for input value and temp is used to compare value.
  • In thew next step, two string variable "SENTINEL and repeat" is declared in which "SENTINEL" is constant and a bool variable "notDone" is declared.
  • After input the value from the user-end a loop is declared that compare and swap value and print its value.

6 0
3 years ago
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